Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        Solution sol = new Solution(N);
        for(int i = 0; i < N; ++i)
            sol.addCoin(i, sc.nextDouble());
        System.out.println(sol.solve());
    }
}

class Solution {
    double[] probs;
    int N;
    double[] dp;
    
    Solution(int N){
        this.N = N;
        this.dp = new double[N+1];
        this.probs = new double[N];
    }
    
    void addCoin(int i, double prob){
        probs[i] = prob;
        if(i == 0){
            dp[1] = probs[0];
            dp[0] = 1 - dp[1];
            return;
        }
        for(int j = N; j >= 1; --j){
            dp[j] = (dp[j]*(1-prob)) + (dp[j-1]*prob);
        }
        dp[0] *= (1 - prob);
    }
    
    double solve() {
        double res = 0;
        for(int i = (N+1)>>1; i <= N; ++i)
            res += dp[i];
        return res;
    }
}